MHT CET · Physics · Center of Mass Momentum and Collision
In the system of two particles of masses ' \(\mathrm{m}_1\) ' and ' \(\mathrm{m}_2\) ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance
- A \(\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}\), towards the centre of mass.
- B \(\frac{m_1}{m_2} d\), away from the centre of mass.
- C \(\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}\), towards the centre of mass.
- D \(\frac{m_2}{m_1} \mathrm{~d}\), away from the centre of mass.
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}\), towards the centre of mass.
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}& m_1 x_1=m_2 x_2 ...(i)\\& m_1\left(x_1-d\right)=m_2\left(x_2-d^{\prime}\right) ...(ii)\\\therefore & m_1 x_1-m_1 d=m_2 x_2-m_2 d^{\prime} \\& m_1 d=m_2 d^{\prime} \\\therefore & d^{\prime}=\frac{m_1}{m_2} d\end{array}\)
...[From (i)]
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